(a) Solution 1. We note that $n=2^{m}$ has $m+1$ divisors, $d_{k}=2^{k-1}$ for $1 \leqslant k \leqslant m+1$. Thus

$$1+d_{1}+\cdots+d_{k-1}=1+\left(2^{k-1}-1\right)=2^{k-1}=d_{k}$$

for each $k \geqslant 2$, and hence each power of 2 is a bad integer. This exhibits infinitely many bad integers.

Remark. It is true more generally that

If $n=2^{r} m$, where $m$ is a product of (odd) primes each less than $2^{r+1}$, then $n$ is bad.

This is an immediate corollary of the previous result and the following observation:

If $n=2^{r} m$ is bad, where $m$ is odd, then so is $p n$ for any odd prime $p$. Let $D_{K}$ be a divisor of $p n$, so $D_{K}=p$ or $D_{K}=d_{k}$ or $D_{K}=p d_{k}$, where $d_{k}>1$ is a divisor of $n$. In the first case, observe that there exists $t<r+1$ such that $2^{t}<p<2^{t+1}$ by assumption. Then $\left\{1,2, \ldots, 2^{t}\right\} \subseteq\left\{D_{1}, \ldots, D_{K-1}\right\}$, and so

$$D_{K}=p<2^{t+1}=1+\left(1+2+\cdots+2^{t}\right) \leqslant 1+D_{1}+\cdots+D_{K-1} .$$

In the final case, $\left\{1,2, \ldots, 2^{t}, p d_{1}, \ldots, p d_{k-1}\right\} \subseteq\left\{D_{1}, \ldots, D_{K-1}\right\}$, and so

$$\begin{aligned} D_{K} & =p d_{k} \leqslant p\left(1+d_{1}+\cdots+d_{k-1}\right)=p+p d_{1}+\cdots+p d_{k} \\ & <2^{t+1}+p d_{1}+\cdots+p d_{k-1}=1+\left(1+\cdots+2^{t}\right)+p d_{1}+\cdots+p d_{k-1}<1+\left(D_{1}+\cdots+D_{K-1}\right) . \end{aligned}$$

In the second case, $\left\{d_{1}, \ldots, d_{k-1}\right\} \subseteq\left\{D_{1}, \ldots, D_{K-1}\right\}$ immediately implies the required inequality, and so $p n$ is indeed bad.

This result is weak, however: only 57931 (6.99\%) of the 829157 bad numbers not larger than $10^{7}$ are of this form.